In $\mathbb{Q}$, $r\sim s$ iff $r-s\in\mathbb{Z}$. Show that this is an equivalence relation and describe the associated partition

Question

Attempt.

Part one

Proof

Define a relation $\sim$ on $\mathbb{Q}$ as $r\sim s$ iff $r-s\in\mathbb{Z}$. To prove that $\sim$ is an equivalence relation, we must verify that $\sim$ is reflexive, symmetric, and transitive.

1) Suppose $r$ is a rational number. Then $r\sim r$ or $r-r=0$ is an integer. So $\sim$ is reflexive.

2) Suppose $r,s$ are rational numbers. Then $r\sim s$ or $r-s$ is an integer. Now, $-(r-s)$ or $s-r$ is also an integer so $s\sim r$. Therefore, $\sim$ is symmetric.

3) Suppose $r,s,t$ are rational numbers. Then $r\sim s$ and $s\sim t$ or $r-s$ is an integer and $s-t$ is an integer. Now, adding $r-s$ and $s-t$ to each other gives $r-t$, which is also an integer and we get $r\sim t$. Therefore, $\sim$ is transitive.

Hence, since $\sim$ is reflexive, symmetric, and transitive, $\sim$ is an equivalence relation.

Part Two

Now, assuming this is correct, I'm having a little difficulty with describing the partition of this equivalence relation. I'd be grateful for any suggestions on how to proceed.

Thanks for your time and attention and I hope this work has not been too painful (I apologize for the poor math and reasoning on my part) too read. Live long and prosper.

Answer 1

Your proof of part one is O.K. ($a,b,c, ...$ are not needed !)

Part two: let us denote the equivalence class of $r$ by $[r]$. Then:

$x \in [r] \iff x \sim r \iff x=r+k $ for some $k \in \mathbb Z$

Hence: $[r]=\{r+k : k \in \mathbb Z\}$

Answer 2

I'm going to try and address the second part in an intuitive way, as I feel this might be helpful in addition to knowing how to shuffle the symbols around properly.

I would start concretely, maybe by asking "what are all the things that are equivalent to 1?" 1 is equivalent to all rational numbers who are an integer distance away from one. So $$\ldots, -2, -1, 0, 1, 2\ldots$$ are all equivalent to 1. If I take something like $\frac{1}{2}$, all the things that are equivalent would be $$\ldots, -1+\frac{1}{2},\ 0+\frac{1}{2},\ 1+\frac{1}{2},\ 2+\frac{1}{2},\ 3+\frac{1}{2},\ldots$$ You can do this starting with any rational number.

What you should notice is that if you take an arbitrary equivalence class, you'll find that the distinguishing feature of that class is the fractional part of all its elements. The integer part doesn't matter, and this is what the relation $r-s\in\mathbf{Z}$ is trying to capture.

Another way to say it is this: We're defining a new type of equality on the rationals. Two rationals are the "same" if they have the same fractional part and they are "different" if their fractional parts are different.